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bdkbZ—14% lglaca/ fo'ys"k.k cuke izrhixeu fo'ys"k.k
14-7 izrhixeu xq.kkadksa }kjk lglEcU/ xq.kkad dk fu/kZj.k (Determination of uksV
Correlation Coefficient By Regression Coefficients)
izrhixeu xq.kkad (Regression Coefficient) og eku n'kkZrk gS tks ,d Js.kh osQ py&ewY;ksa esa bdkbZ
ifjorZu (unit change) gksus ls nwljh Js.kh osQ py&ewY;ksa esa vkSlru ifjorZu gksxkA ;g izrhixeu js[kkvksa
osQ lEcU/ esa js[kk <yku (slope) dk chtxf.krh; eki gSA tc nks izrhixeu xq.kkad Kkr gksa rks budh lgk;rk
ls lglEcU/ xq.kkad (r) Kkr fd;k tk ldrk gSA lglEcU/ xq.kkad (r) nksuksa izrhixeu xq.kkadksa (b ,oa b )
xy yx
osQ xq.kuiQyksa dk oxZewy gksrk gS vFkkZr~ lglEcU/ xq.kkad nks izrhixeu xq.kkadksa dk xq.kksÙkj ekè; gksrk gSA
σ σ y
2
b xy × b yx = r σ x y × r σ x = r = r
fVIi.khμ
(1) è;ku jgs fd b , b rFkk r dk fpÉ leku jgrk gSA rhuksa /ukRed gksaxs ;k rhuksa Í.kkRed gksaxsA
xy yx
(2) nksuksa izrhixeu xq.kkad lkekU;r% ,d ls vf/d ewY; osQ ugha gks ldrsA ;fn b vkSj b nksuksa dk ewY;
xy yx
2
1 ls vf/d gksxk rks nksuksa dk xq.kuiQy r Hkh 1 ls vf/d gksxk ftlosQ iQyLo:i r Hkh 1 ls vf/d
gksxk tks vlEHko gSA gka] ;g lEHko gS fd ,d xq.kkad dk eku 1 ls vf/d gks ldrk gS] ijUrq bl
voLFkk esa nwljs xq.kkad dk eku bruk de gksuk pkfg, fd nksuksa dk vkil esa xq.kk djus ij xq.kuiQy
1 ls vf/d u gksA
mnkgj.k (Illustration) 6
(i) ;fn b = 0.64 rFkk b = 1 gks rks r dk eku crkb,A
yx xy
Find out the value of r if b = 0.64 and b = 1.
yx xy
(ii) ;fn nks izrhixeu xq.kkad – 0.9 rFkk – 0.5 gksa rks lglEcU/ xq.kkad dk eku crkb,A
If the two regression coefficient are – 0.9 and – 0.5, find out the value of correlation
coefficient.
gy (Solution)
(i) b = .64, b = 1
yx xy
∴ r = b yx × b xy = × .64 1 = . 0 64 = 0.8
(ii) ekuk b = – 0.9 rFkk b = – 0.5
yx xy
.
rc r = – 09. × 0 5 = . − 0 45 = – .671
Lo&ewY;kadu (Salf Assessment)
1- lgh fodYi pqfu,
1. ;fn nks pjksa ds chp lglaca/ ik;k tkrk gS ijarq dkj.k ifj.kke laca/ ugha ik;k tkrk rks
ml lglaca/ dks dgk tkrk gS&
(d) izR;{k laca/ ([k) fujFkZd laca/
(x) viR;{k laca/ (?k) buesa ls dksbZ ughaA
LOVELY PROFESSIONAL UNIVERSITY 231